How do you identify all asymptotes or holes for #f(x)=(x^2-16)/(x^2-5x+4)#?

Question

Anna Allen · Accepted Answer

hole at x = 4
vertical asymptote at x = 1
horizontal asymptote at y = 1

Factorise and simplify f(x) #f(x)=(cancel((x-4))(x+4))/(cancel((x-4))(x-1))=(x+4)/(x-1)# We have removed a duplicate factor from the numerator/denominator of f(x). This indicates there is a hole at x = 4 This hole is not in the simplified version of f(x). The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote. solve : #x-1=0rArrx=1" is the asymptote"# Horizontal asymptotes occur as #lim_(xto+-oo),f(x)toc" ( a constant)"# divide terms on numerator/denominator by x #f(x)=(x/x+4/x)/(x/x-1/x)=(1+4/x)/(1-1/x)# as #xto+-oo,f(x)to(1+0)/(1-0)# #rArry=1" is the asymptote"# graph{(x+4)/(x-1) [-10, 10, -5, 5]}

Caleb Alexander · Answer

undefined To identify all asymptotes or holes for $ f(x) = \frac{x^2 - 16}{x^2 - 5x + 4} $, we first need to analyze the behavior of the function as $ x $ approaches certain values.

1. Vertical Asymptotes:
   Vertical asymptotes occur where the denominator of the function approaches zero while the numerator does not. Set the denominator equal to zero and solve for $ x $ to find potential vertical asymptotes. In this case, the denominator $ x^2 - 5x + 4 $ factors as $ (x - 1)(x - 4) $, so vertical asymptotes occur at $ x = 1 $ and $ x = 4 $.

2. Holes:
   Holes occur where there is a factor common to both the numerator and denominator that cancels out. Factor both the numerator and denominator and simplify. If any factors cancel out, you have found a hole. In this case, $ x^2 - 16 $ factors as $ (x - 4)(x + 4) $, so there's a common factor of $ (x - 4) $. After canceling out this common factor, we find that there is a hole at $ x = 4 $.

3. Horizontal Asymptotes:
   Horizontal asymptotes occur as $ x $ approaches positive or negative infinity. To find horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is at $ y = 0 $. If the degrees are equal, divide the leading coefficients to find the horizontal asymptote. If the degree of the numerator is greater, there is no horizontal asymptote. In this case, both the numerator and denominator have the same degree (2), so we divide the leading coefficients to find the horizontal asymptote.

Therefore, the asymptotes and holes for the function $ f(x) = \frac{x^2 - 16}{x^2 - 5x + 4} $ are:
- Vertical asymptotes at $ x = 1 $ and $ x = 4 $
- Hole at $ x = 4 $
- No horizontal asymptote.